LeetCode:House Robber III

House Robber III

Total Accepted: 13977 Total Submissions: 37027 Difficulty: Medium

The thief has found himself a new place for his thievery again. There is only one entrance to this area, called the "root." Besides the root, each house has one and only one parent house. 

After a tour, the smart thief realized that "all houses in this place forms a binary tree". It will automatically contact the police if two directly-linked houses were broken into on the same night.

Determine the maximum amount of money the thief can rob tonight without alerting the police.

Example 1:

     3    / \   2   3    \   \      3   1

Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.

Example 2:

     3    / \   4   5  / \   \  1   3   1

Maximum amount of money the thief can rob = 4 + 5 = 9.

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

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 (E) House Robber (M) House Robber II

思路：

原问题分为：1）偷当前结点；2）不偷当前结点；两个问题

1）时的amount 为4个“孙子”结点的值的和；

2）时的amount为2个“孩子”结点的值的和。

java code:

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public int rob(TreeNode root) {                if(root == null) return 0;                int valOfNoRoot = 0; // 不偷当前结点        int valOfRoot = 0; // 偷当前结点                valOfNoRoot = rob(root.left) + rob(root.right);                if(root.left != null)            valOfRoot += rob(root.left.left) + rob(root.left.right);        if(root.right != null)            valOfRoot += rob(root.right.left) + rob(root.right.right);                return Math.max(valOfRoot + root.val, valOfNoRoot);            }}

这段代码花了：1133 ms

java代码勉强可以通过，但是c++过不了（c++的时间要求就短一些）。

这是由于递归过程中有大量的重复计算，因此可以将计算过的值保存起来，减少计算次数。

使用HashMap来保存已计算值的java code:

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {     public int rob(TreeNode root) {        HashMap<TreeNode, Integer> map = new HashMap<TreeNode, Integer>();        return rob(root, map);    }        private int rob(TreeNode root, HashMap<TreeNode, Integer> map) {        if(root == null) return 0;                if(map.containsKey(root)) return map.get(root);                int valOfNoRoot = 0; // 不偷当前结点        int valOfRoot = 0; // 偷当前结点                valOfNoRoot = rob(root.left, map) + rob(root.right, map);                if(root.left != null)            valOfRoot += rob(root.left.left, map) + rob(root.left.right, map);        if(root.right != null)            valOfRoot += rob(root.right.left, map) + rob(root.right.right, map);                int val = Math.max(valOfRoot + root.val, valOfNoRoot);        map.put(root, val);        return val;    }}

时间缩短到了：8 ms

c++ code:

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    int rob(TreeNode* root) {                vector<int> ret = robSub(root);        return max(ret[0], ret[1]);    }        vector<int> robSub(TreeNode* root) {                if(!root) return vector<int>(2);                vector<int> left= robSub(root->left);        vector<int> right= robSub(root->right);                vector<int> ret(2);        ret[0] = max(left[0], left[1]) + max(right[0], right[1]); // 不偷当前结点        ret[1] = root->val + left[0] + right[0]; // 偷当前结点                return ret;    }};